Question

A department store manager to estimate at a %90 confidence level the mean about by all customers at this store

Answer 1

The answer is "287.19"

Explanation:

Please find the complete question in the attached file.

`\alpha=1-0.90=0.10\\\\(\alpha)/(2)=0.05\\\\Z=1.64\\\\Marginal \ error=Z * (\sigma_x)/(√(n))\\\\n=((Z * \sigma_x)/(error))^2\\\\n=((1.64 * 31)/(3))^2\\\\n=287.1895\approx 287.19`